Moment of inertia problem involving a cylinder rolling down an incline

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Homework Help Overview

The discussion revolves around a moment of inertia problem involving a cylinder rolling down an incline. Participants are examining the application of energy conservation principles and the calculations related to potential and kinetic energy in this context.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the final velocity and moment of inertia using energy conservation but questions the validity of their approach and calculations.
  • Some participants question the assumptions made about the cylinder's uniformity and the interpretation of the height used in calculations.
  • Others raise concerns about unit conversions and the specific values used in the potential energy calculations.
  • There are discussions about the effects of temperature on energy conservation in practical demonstrations.

Discussion Status

Participants are actively engaging in clarifying assumptions and calculations. Some guidance has been offered regarding the interpretation of the height and the uniformity of the cylinder, but no consensus has been reached on the correct approach to the problem.

Contextual Notes

There are constraints regarding the provided information, such as the height of the incline and the mass units. Participants are also exploring the implications of temperature on energy conservation in practical scenarios.

as2528
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Homework Statement
A metal can containing condensed mushroom soup has a mass of 215 g, a height of 10.8 cm, and a diameter of 6.38 cm. It is placed at rest on its side at the top of a 3.00-m-long incline that is at an angle of 25.0° to the horizontal and is then released to roll straight down. Assuming energy conservation, calculate the moment of inertia of the can if it takes 1.50 s to reach the bottom
of the incline. Which pieces of data, if any, are unnecessary for calculating the solution?
Relevant Equations
vf=vi+at
K=1/2*I*w^2+1/2*m*v^2
PE=mgh
v=rw
a=2/3*g*sin(B)
a=2/3*g*sin(25*(pi/180))=>a=2.8507 m/s^2
vf=vi+at=>vf=0+2.8507*1.50=>vf=4.2760 m/s

So the translational motion of the cylinder is 4.2760 m/s.

4.2760=R*w
w=134.04 rad/s
PE=mgh=>PE=215*9.8*.108=>PE=227.56 J
PE = KE at the end of the roll because of energy conservation.

227.56 = 1/2*I*w^2+1/2*m*v^2
227.56=0.5*I*(134.04)^2+0.5*215*(4.276)^2
I=-0.1931 kg*m^2

The answer is 1.21 *10^4 kg*m2

How is this solved and why is my approach wrong? I think I did all the calculations correctly, so I must have gone wrong in applying the physics.
 
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as2528 said:
Relevant Equations::

PE=mgh
When calculating the PE, careful with the units. The mass is given in grams.

as2528 said:
a=2/3*g*sin(B)
You shouldn't assume that the can of soup is a uniform solid cylinder. So, this formula might not apply. You are given enough information to find the acceleration without this assumption.

The answer is 1.21 *10^4 kg*m2
Looks like you dropped the negative sign for the power of 10.
 
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As an aside, I should add that the conservation of energy in this situation is temperature-dependent. I used to do a lecture demonstration in which two cans of the same soup (chunky beef) raced down an incline. One was at room temperature, the other had spent the night in the freezer and was still frozen. Guess which can won the race.
 
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kuruman said:
As an aside, I should add that the conservation of energy in this situation is temperature-dependent. I used to do a lecture demonstration in which two cans of the same soup (chunky beef) raced down an incline. One was at room temperature, the other had spent the night in the freezer and was still frozen. Guess which can won the race.
Try a similar demonstration with a raw egg versus a hard-boiled egg.
 
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as2528 said:
PE=215*9.8*.108
In addition to the mass units error, where does the .108 come from? It is not 3sin(25°).
 
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kuruman said:
As an aside, I should add that the conservation of energy in this situation is temperature-dependent. I used to do a lecture demonstration in which two cans of the same soup (chunky beef) raced down an incline. One was at room temperature, the other had spent the night in the freezer and was still frozen. Guess which can won the race.
Is that to do with conservation of energy or just the fact that the contents are not rotating as fast?
 
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haruspex said:
In addition to the mass units error, where does the .108 come from? It is not 3sin(25°).
The .108 was supposed to be the height of the incline.
 
as2528 said:
The .108 was supposed to be the height of the incline.
3m at 25°, only 11cm? Looks like you used 2°.
 
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haruspex said:
3m at 25°, only 11cm? Looks like you used 2°.
No I used it from the question. It said that the height was 10.8 cm. I converted that to meters and reasoned that since gravity is conservative I could use .108 as the height of the soup can.
 
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as2528 said:
It said that the height was 10.8 cm.
That is the height of the can as a cylinder, or length if that's clearer, not the height at which the can is placed.
 
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  • #11
haruspex said:
That is the height of the can as a cylinder, or length if that's clearer, not the height at which the can is placed.
Oh! I did not realize that. Thanks!
 
  • #12
haruspex said:
Is that to do with conservation of energy or just the fact that the contents are not rotating as fast?
The latter. I believe that the energy goes into viscous losses between the solid tidbits in the can and the broth plus the coupling between the fluid inside and the walls.

On edit:
Here is a homemade video of a can of tomato sauce given an impulse on my kitchen counter. What happens to the initial energy?

 
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